# Computing the matrix of the linear map $X\rightsquigarrow AXA^t$, where $A=\left[\begin{smallmatrix}2&1\\0&1\end{smallmatrix}\right]$

Let $V$ be the vector space of real $2\times 2$ symmetric matrices $X=\begin{bmatrix}x&y\\y&z\end{bmatrix}$ and let $A=\begin{bmatrix}2&1\\0&1\end{bmatrix}$. Determine the matrix of the linear operator on $V$ defined by $X\rightsquigarrow AXA^t$ with respect to a suitable basis.

So far, I got $AXA^t=\begin{bmatrix}4x+4y+z&2y+z\\2y+z&z\end{bmatrix}$

I need to find a matrix $C$ such that $CX=AXA^t$, but not find $C$. Am I doing anything wrong. Please someone tell me what is wrong here.

• I think you have a small misunderstanding about the property that $C$ should satisfy. Letting $C$ be the matrix of your linear operator, then $C$ satisfies $C[X]_\mathcal{B} = [AXA^t]_\mathcal{B}$, where $[X]_\mathcal{B}$ is the coordinate vector of $X$ with respect to a basis $\mathcal{B}$ for $V$. Since $V$ is $2$-dimensional (Why? Can you find a basis for $V$?), then $[X]_\mathcal{B} \in \mathbb{R}^2$ for each $X \in V$, and $C$ is a $2 \times 2$ matrix. – André 3000 Oct 21 '17 at 13:34
• @Quasicoherent $V$ is $3$-dimension, isn't? Basis is $\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix} \}$. – algebra_001 Oct 21 '17 at 13:57
• Whoops, you're right: my mistake! Sorry, I missed that $x$ and $z$ were different. The rest of my comment is still true, though, except that $[X]_\mathcal{B} \in \mathbb{R}^3$ and $C$ should be $3 \times 3$. – André 3000 Oct 21 '17 at 14:11
• @Quasicoherent , it just hit me hard!!! Using the basis I wrote, the co-ordinate of V should be $(x,y,z)$, isn't? But $(x,y,z)$ is not even a element of $V$.... how is that possible! – algebra_001 Oct 21 '17 at 17:01

First, some generalities. The structure of a vector space is totally determined by its dimension. Given a real vector space $V$ of dimension $n$, then $V$ is isomorphic to $\mathbb{R}^n$, where the isomorphism is given by any choice of basis for $V$. More explicitly, given a basis $\newcommand{\B}{\mathcal{B}} \newcommand{\R}{\mathbb{R}} \B = \{v_1, \ldots, v_n\}$ for $V$, we get an isomorphism given by the coordinate map \begin{align*} \varphi_\B: V &\to \R^n\\ x &\mapsto [x]_\B \end{align*} where $[x]_\B \in \R^n$ is the coordinate vector of the vector $x \in V$, whose entries are the unique weights $c_1, \ldots, c_n \in \R$ expressing $x$ as a linear combination of $v_1, \ldots, v_n$. That is, $$[x]_\B = \begin{pmatrix} c_1\\ \vdots\\ c_n \end{pmatrix}$$ where $$x = c_1 v_1 + \cdots + c_n v_n \, .$$ Once we've chosen a basis and identified with $\R^n$, a linear map $T: V \to V$ can be expressed as left-multiplication by some matrix $[T]_\B$. This is captured by the following commutative diagram. $\hspace{6.5cm}$ As for computing $[T]_\B$, one can show that $$[T]_\B = \begin{pmatrix} | & & |\\ [T(v_1)]_\B & \cdots & [T(v_n)]_\B\\ | & & | \end{pmatrix}$$ where as before $\B = \{v_1, \ldots, v_n\}$.
Returning to your problem, you have found a basis for $V$, and also determined the general formula for $T$. Now you just have to apply $T$ to the basis vectors, write each image as a linear combination of the basis vectors, and use these as the columns of $[T]_\B$. For instance, your first basis vector is $v_1 = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$ and $$T(v_1) = \begin{pmatrix}4 & 0\\ 0 & 0\end{pmatrix} = 4 v_1$$ so $[T(v_1)]_\B = \begin{pmatrix}4\\ 0\\ 0\end{pmatrix}$. Can you compute the remaining columns?
• Yeah! I actually got the answer. I found $$\begin{bmatrix}4&0&0\\4&2&0\\1&1&1\end{bmatrix}$$ – algebra_001 Oct 23 '17 at 4:28